If $A$ is positive definite, then $\int_{\mathbb{R}^n}\mathrm{e}^{-\langle Ax,x\rangle}\text{d}x=\left|\det\left({\pi}^{-1}A\right)\right|^{-1/2}$

Solution 1:

As the matrix $A$ is symmetric (a positive definite is by definition also symmetric), then it is also diagonalizable, i.e., $A=U^TDU$, where $U$ is orthogonal ($U^TU=UU^T=I$ and $|\det U|=1$) and $$ D=\mathrm{diag}(\lambda_1,\ldots,\lambda_n), $$ where $\lambda_i>0$, for $i=1,\ldots,n$.

Clearly $\det A=\prod_{i=1}^n \lambda_i$. First, we have that $$ \langle x,Ax\rangle=\langle x,U^TDUx\rangle=\langle Ux,DUx\rangle, $$ and hence $$ \int_{\mathbb R^n}\exp(-\langle x,Ax\rangle)\,dx =\int_{\mathbb R^n}\exp(-\langle Ux,DUx\rangle)\,dx. $$ The Theorem of Integration by Change of Variables provides that \begin{align} \int_{\mathbb R^n}\exp(-\langle Ux,DUx\rangle)\,dx\stackrel{y=Ux}=&\int_{\mathbb R^n}\exp(-\langle y,Dy\rangle)\,|\det U|\,dy \\ =&\int_{\mathbb R^n}\exp(-\langle y,Dy\rangle)\,dy\\ =&\int_{\mathbb R^n} \exp(-(\lambda_1y_1^2+\lambda_2y_2^2+\cdots+\lambda_ny_n^2)\rangle)\,dy_1\,dy_2\cdots dy_n \\ \stackrel{(*)}{=}&\prod_{i=1}\int_{\mathbb R}\exp(-\lambda_iy_i^2)\,dy_i\stackrel{(\dagger)}{=}\prod_{i=1}^n\left(\frac{\pi}{\lambda_i}\right)^{1/2}=\pi^{n/2}|\det A|^{-1/2} \\=&\big|\det(\pi^{-1}A)\big|^{-1/2}, \end{align} where at $(\star)$ we used Fubini's Theorem, while at $(\dagger)$ the fact that $$ \int_{\mathbb R}\mathrm{e}^{-\lambda x^2}=\sqrt{\frac{\pi}{\lambda}}, $$ whenever $\lambda>0$.

Solution 2:

Assume first that $A$ is diagonal. Then the result follows from Fubini's theorem and the equality $$\int_{\mathbb R}e^{-x^2/2}\mathrm dx=\sqrt{2\pi}.$$

Now use spectral theorem to write $A=P^tDP$, where $D$ is diagonal with positive diagonal entries and $P$ orthogonal.